Question

Which from the following equations represents the relation between solubility ( $\text{mol L}^{-1}$ ) and solubility product for the salt $\text{B}_2\text{A}$ ?

• A. $\text{S} = \left( \frac{\text{K}_{\text{sp}}}{4} \right)^{1/3}$
• B. $\text{S} = (4 \text{K}_{\text{sp}})^{1/3}$
• C. $\text{S} = \left( \frac{\text{K}_{\text{sp}}}{3} \right)^{1/3}$
• D. $\text{S} = (3 \text{K}_{\text{sp}})^{1/3}$

Hint:

Shortcut: For $A_xB_y$: $K_{sp} = x^x y^y S^{x+y}$

Answer 1

The Correct Option is A

Concept: For salt $\text{B}_2\text{A}$: \[ \text{B}_2\text{A} \rightleftharpoons 2\text{B}^+ + \text{A}^{2-} \]

Step 1: Assume solubility = S.
• $[\text{B}^+] = 2S$
• $[\text{A}^{2-}] = S$

Step 2: Write $K_{sp}$ expression. \[ K_{sp} = [\text{B}^+]^2 [\text{A}^{2-}] \] \[ K_{sp} = (2S)^2 \times S = 4S^3 \]

Step 3: Solve for S. \[ S^3 = \frac{K_{sp}}{4} \Rightarrow S = \left(\frac{K_{sp}}{4}\right)^{1/3} \]

Step 4: Conclusion.
Thus, correct relation is $\left(\frac{K_{sp}}{4}\right)^{1/3}$. Final Answer: Option (A)